suppose the distance(in feet) that an object travels in t seconds is given by the formula s(t)= 2t^3 + 45 - 5. find s(2), v(2) and a(2)
To find s(2), v(2), and a(2) from the given formula, we need to differentiate the position function s(t) with respect to time. First, let's find s(2).
Given: s(t) = 2t^3 + 45t - 5
To find s(2), we substitute t = 2 into the formula:
s(2) = 2(2)^3 + 45(2) - 5
= 16 + 90 - 5
= 101 feet
Therefore, s(2) = 101 feet.
Now, let's find v(2) by differentiating s(t) with respect to t (using the power rule for differentiation):
v(t) = d/dt [2t^3 + 45t - 5]
v(t) = 6t^2 + 45
Now, substitute t = 2:
v(2) = 6(2)^2 + 45
= 6(4) + 45
= 24 + 45
= 69 feet/second
Therefore, v(2) = 69 feet/second.
To find a(2), differentiate v(t) = 6t^2 + 45 with respect to t again:
a(t) = d/dt [6t^2 + 45]
a(t) = 12t
Now, substitute t = 2:
a(2) = 12(2)
= 24 feet/second^2
Therefore, a(2) = 24 feet/second^2.
To summarize:
s(2) = 101 feet
v(2) = 69 feet/second
a(2) = 24 feet/second^2
do ur hw scrub
v(t) = 6t^2 + ?
a(t) = 12t + ?
The ? is because you did not specify the power of t for the "45" term
Anyway, fill it out and use your numbers.
scrub...